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| learn:courses:real-analog-chapter-8:start [2017/05/01 22:25] – [Section Summary] Martha | learn:courses:real-analog-chapter-8:start [2022/10/04 11:38] (current) – external edit 127.0.0.1 | ||
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| ====== Real Analog: Chapter 8 ====== | ====== Real Analog: Chapter 8 ====== | ||
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| + | * Chapter 8 Video: | ||
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| ====== 8. Introduction and Chapter Objectives ====== | ====== 8. Introduction and Chapter Objectives ====== | ||
| - | Second order systems are, by definition, systems whose input-output relationship is a //second order// differential equation. A second order differential equation contains a //second order derivative// | + | Second order systems are, by definition, systems whose input-output relationship is a //second order// differential equation. A second order differential equation contains a //second order derivative// |
| Second order systems, like first order systems, are an extremely important class of systems. In previous chapters, we saw that the natural response of first order systems decays exponentially with time – the natural response decays monotonically to zero. The natural response of second order systems can, however, oscillate with time – we will see that a second order systems response can contain sinusoidal components. The motion of a pendulum, for example, can be modeled by a second order system. These oscillations are due to the transfer of energy between the two energy storage mechanisms; a pendulum, for example, oscillates because of the cyclic exchange of potential and kinetic energy of the mass. | Second order systems, like first order systems, are an extremely important class of systems. In previous chapters, we saw that the natural response of first order systems decays exponentially with time – the natural response decays monotonically to zero. The natural response of second order systems can, however, oscillate with time – we will see that a second order systems response can contain sinusoidal components. The motion of a pendulum, for example, can be modeled by a second order system. These oscillations are due to the transfer of energy between the two energy storage mechanisms; a pendulum, for example, oscillates because of the cyclic exchange of potential and kinetic energy of the mass. | ||
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| $$\omega = \frac{2\pi}{T} | $$\omega = \frac{2\pi}{T} | ||
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| === Note: === | === Note: === | ||
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| $$A=|A| e^{j \theta_A} | $$A=|A| e^{j \theta_A} | ||
| - | We can generalize our definition of the complex exponential to time-varying signals. If we define a time varying signal $e^{j \omega t}$$ , we can use equation (8.31) to write: | + | We can generalize our definition of the complex exponential to time-varying signals. If we define a time varying signal $e^{j \omega t}$ , we can use equation (8.31) to write: |
| $$e^{\pm j \omega t} = \cos \omega t \pm j \sin \omega t (Eq. 8.33)$$ | $$e^{\pm j \omega t} = \cos \omega t \pm j \sin \omega t (Eq. 8.33)$$ | ||
| - | The signal $e^{j \omega t}$ can be visualized as a unit vector rotating around the origin in the complex plane; the tip of the vector scribes a unit circle with its center at the origin of the complex plane. This is illustrated in Figure 8.7. The vector rotates at a rate defined by the quantity $\omega$ -- the vector makes one complete revolution every $\frac{2\pi}{\omega} seconds. | + | The signal $e^{j \omega t}$ can be visualized as a unit vector rotating around the origin in the complex plane; the tip of the vector scribes a unit circle with its center at the origin of the complex plane. This is illustrated in Figure 8.7. The vector rotates at a rate defined by the quantity $\omega$ -- the vector makes one complete revolution every $\frac{2\pi}{\omega}$ seconds. |
| The projection of this rotating vector on the real axis traces out the signal $\cos \omega t$ , as shown in Figure 8.7, while the projection of the rotating vector on the imaginary axis traces out the signal $\sin \omega t$, also shown in Figure 8.7. | The projection of this rotating vector on the real axis traces out the signal $\cos \omega t$ , as shown in Figure 8.7, while the projection of the rotating vector on the imaginary axis traces out the signal $\sin \omega t$, also shown in Figure 8.7. | ||
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| In equation (8.37), the $e^{-\zeta \omega_n t}$ term is a decaying exponential with time constant $\frac{1}{\zeta \omega_n}$. The $e^{\omega_n t \sqrt{\zeta^2-1}}$ is a growing exponential with time constant $\frac{1}{\omega_n \sqrt{\zeta^2-1}}$. Thus, the overall system response is a sum of two decaying exponential signals, one which is proportional to $e^{-\zeta \omega_n t} \cdot e^{\omega_n t \sqrt{\zeta^2-1}}$ and the other which is proportional to $e^{-\zeta \omega_n t} \cdot e^{-\omega_n t \sqrt{\zeta^2-1}}$. | In equation (8.37), the $e^{-\zeta \omega_n t}$ term is a decaying exponential with time constant $\frac{1}{\zeta \omega_n}$. The $e^{\omega_n t \sqrt{\zeta^2-1}}$ is a growing exponential with time constant $\frac{1}{\omega_n \sqrt{\zeta^2-1}}$. Thus, the overall system response is a sum of two decaying exponential signals, one which is proportional to $e^{-\zeta \omega_n t} \cdot e^{\omega_n t \sqrt{\zeta^2-1}}$ and the other which is proportional to $e^{-\zeta \omega_n t} \cdot e^{-\omega_n t \sqrt{\zeta^2-1}}$. | ||
| - | The term $e^{-\zeta \omega_n t} \cdot e^{\omega_n t \sqrt{\zeta^2-1}}$ is the product of two exponentials: | + | The term $e^{-\zeta \omega_n t} \cdot e^{\omega_n t \sqrt{\zeta^2-1}}$ is the product of two exponentials: |
| - | The term $e^{-\zeta \omega_n t} \cdot e^{-\omega_n t \sqrt{\zeta^2-1}}$ is the product of two exponentials: | ||
| The term $e^{\zeta \omega_n t} \cdot e^{-\omega_n t \sqrt{\zeta^2-1}}$ is the product of two decaying exponentials; | The term $e^{\zeta \omega_n t} \cdot e^{-\omega_n t \sqrt{\zeta^2-1}}$ is the product of two decaying exponentials; | ||
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| **d.)** In order for the system to be critically damped, the damping ratio $\zeta = 1$. From equation (8.47) with $\zeta = 1$, we obtain: | **d.)** In order for the system to be critically damped, the damping ratio $\zeta = 1$. From equation (8.47) with $\zeta = 1$, we obtain: | ||
| - | $$2 \zeta \omega_n = \frac{1R}{L} \Rightarrow 2(1)(1000) = \frac{2R}{1H} \Rightarrow = 1000 \Omega$$ | + | $$2 \zeta \omega_n = \frac{1R}{L} \Rightarrow 2(1)(1000) = \frac{2R}{1H} \Rightarrow |
| **e.)** Initial conditions on $v_C(t)$ are $v_C(0)$ and $\frac{dv_C(t)}{dt} |_{t=0}$ We are given $v_C(-)=1V$ in the problem statement, but we need to determine $\frac{dv_C(t)}{dt} |_{t=0}$; the current through the inductor, $i_L(0)$ can be used to determine this. The current through the inductor is related to the capacitor voltage via equation (8.43) above: | **e.)** Initial conditions on $v_C(t)$ are $v_C(0)$ and $\frac{dv_C(t)}{dt} |_{t=0}$ We are given $v_C(-)=1V$ in the problem statement, but we need to determine $\frac{dv_C(t)}{dt} |_{t=0}$; the current through the inductor, $i_L(0)$ can be used to determine this. The current through the inductor is related to the capacitor voltage via equation (8.43) above: | ||
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| $$y(t)=y_h(t) + y_p(t)$$ | $$y(t)=y_h(t) + y_p(t)$$ | ||
| - | The homogeneous solution of second order differential equations has been discussed in sections 8.1 and 8.4 and will not be repeated here. The particular solution of the differential equation (8.51) can be obtained by examining the solution to the equation after the homogeneous solution has died out. Letting $t \rightarrow \infty$ in equation (8.51) and noting that the forcing function is a constant as $t \rightarrow \infty$ allows us to set $\frac{d^2y(t \rightarrow \infty}{dt^2} = \frac{dy(t \rightarrow \infty)}{dt} = 0$ amd thus, | + | The homogeneous solution of second order differential equations has been discussed in sections 8.1 and 8.4 and will not be repeated here. The particular solution of the differential equation (8.51) can be obtained by examining the solution to the equation after the homogeneous solution has died out. Letting $t \rightarrow \infty$ in equation (8.51) and noting that the forcing function is a constant as $t \rightarrow \infty$ allows us to set $\frac{d^2y(t \rightarrow \infty)}{dt^2} = \frac{dy(t \rightarrow \infty)}{dt} = 0$ and thus, |
| $$\omega_{n}^{2} y_P(t) = A \Rightarrow y_P(t) = \frac{A}{\omega_{n}^{2}} | $$\omega_{n}^{2} y_P(t) = A \Rightarrow y_P(t) = \frac{A}{\omega_{n}^{2}} | ||
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